Since, $U : H \to H$ satisfies $\lVert U \rVert \le 1$, then $$\left<(IU)h,h\right> = \lVert h \rVert^2  \left< Uh,h \right> \ge \lVert h \rVert^2 (1  \lVert U \rVert) \ge 0 \text{ for all } h \in H$$ We claim that, $N(IU) = R(IU)^{\perp}$
If, $h \in N(IU)$ then, $\left<(IU)(h  th'), h  th'\right> \ge 0 \implies t^2\left<(IU)h' , h'\right>  t\left<(IU)h' , h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$.
It follows that $\left<(IU)h', h\right> = 0$ for all $h' \in H$, i.e., $N(IU) \subseteq R(IU)^{\perp}$.
To see the other way inclusion, similarly if $h \in R(IU)^{\perp}$, then $\left<(IU)(th'  h),th'  h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$,
Implies $t^2\left<(IU)h' ,h'\right>  t\left<(IU) h,h'\right> \ge 0$ for all $t \in \mathbb{R}$.
Hence, $\left<(IU) h,h'\right> = 0$ for all $h' \in H$. That is $h \in N(IU)$.
Now, since $N((IU)^{*}) = R(IU)^{\perp}$ (this is standard result for adjoint of an operator between Banach spaces), it follows $N(IU) = N(I  U^{*})$, i.e., $U(h) = h \iff U^{*}(h) = h$.
